# Probability with inequality condition

Can someone explain how to solve this problem. Since I can get 7 integers from 1st inequality 5 integers from 2nd inequality I got total number of cases of 35. Then I counted the possibilities that would be less than 4. For example x=o y can = 0, 1, 2, 3, when x=1 y=0,1,2 and so on. I came down to 10/35, but the answer is 1/3 .

• This problem isn't very clear. Are $x$ and $y$ supposed to be random variables? If so, what are their distributions? – Math1000 Dec 6 '14 at 2:34
• Sorry, that was all I was given. I believe they don't have any distribution and that they're integers. – GMATnoob Dec 6 '14 at 2:37
• Can you first confirm that $x$ and $y$ are integers, not just real numbers? This makes a difference. – peterwhy Dec 6 '14 at 2:42
• The answer is $\frac{1}{3}$ if they're real numbers with uniform distribution. (A four by six rectangle has area 24, the right triangle cut off by $x+y < 4$ and $x,y \geq 0$, with side lengths four and four, has area 8.) – aes Dec 6 '14 at 2:42
• Confirmed they are real numbers with uniform dist. aes, could you explain further? not sure how you got side lengths four and four. – GMATnoob Dec 6 '14 at 3:03

Here's a six by four rectangle, representing the possible values of $x$ and $y$. Choosing $x$ and $y$ uniformly and independently means choosing a random point in this rectangle, with the probability it's in a given region proportional to the area of that region.
I've drawn a line $x+y = 4$. The values you're interested in are below that line.
The total area is 24, and the area of the triangle below $x+y=4$ is $8$.
Thus the probability that $x+y < 4$ is $\frac{8}{24} = \frac{1}{3}$.
• @GMATnoob Certainly. Also, you could look at $x+y=4$ differently: both $(4,0)$ and $(0,4)$ are solutions, and it's linear, so the line is the line through those two points. – aes Dec 6 '14 at 3:47